JEE Challenger
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ChemMedMain 2026 • 2nd Apr - Shift 1

Compare Bond Lengths and Unpaired Electrons in Dioxygen Species

Given below are two statements : Statement (I) : The correct sequence of bond lengths in the following species is : O2+<O2<O2−<O22−\text{O}_2^+ < \text{O}_2 < \text{O}_2^- < \text{O}_2^{2-} Statement (II) : The correct sequence of number of unpaired electrons in the following species is : O2>O2+>O2−>O22−\text{O}_2 > \text{O}_2^+ > \text{O}_2^- > \text{O}_2^{2-} In the light of the above statements, choose the correct answer from the options given below :

Options

A

Both Statement I and Statement II are true

B

Both Statement I and Statement II are false

C

Statement I is true but Statement II is false

Correct
D

Statement I is false but Statement II is true

Topics & Concepts

Chemical Bonding and Molecular StructureMolecular Orbital Theory

Step-by-Step Solution

To determine the correctness of Statement (I) and Statement (II), we utilize Molecular Orbital Theory (MOT) for dioxygen species.

1. Analysis of Molecular Orbitals, Bond Orders, and Bond Lengths (Statement I):

The valence molecular orbital configuration for the dioxygen species (16-electron system and its ions) is: σ1s2σ1s∗2σ2s2σ2s∗2σ2pz2(π2px2=π2py2)(π2px∗1=π2py∗1)\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \left(\pi_{2p_x}^2 = \pi_{2p_y}^2\right) \left(\pi_{2p_x}^{*1} = \pi_{2p_y}^{*1}\right)

The Bond Order (B.O.\text{B.O.}) is calculated as: B.O.=Nb−Na2\text{B.O.} = \frac{N_b - N_a}{2} where NbN_b is the number of bonding electrons and NaN_a is the number of antibonding electrons.

  • O2+\text{O}_2^+ (15 electrons):

    • Nb=10,Na=5N_b = 10, N_a = 5
    • B.O.=10−52=2.5\text{B.O.} = \frac{10 - 5}{2} = 2.5

  • O2\text{O}_2 (16 electrons):

    • Nb=10,Na=6N_b = 10, N_a = 6
    • B.O.=10−62=2.0\text{B.O.} = \frac{10 - 6}{2} = 2.0

  • O2−\text{O}_2^- (17 electrons):

    • Nb=10,Na=7N_b = 10, N_a = 7
    • B.O.=10−72=1.5\text{B.O.} = \frac{10 - 7}{2} = 1.5

  • O22−\text{O}_2^{2-} (18 electrons):

    • Nb=10,Na=8N_b = 10, N_a = 8
    • B.O.=10−82=1.0\text{B.O.} = \frac{10 - 8}{2} = 1.0

Since Bond Length is inversely proportional to Bond Order (Bond Length∝1B.O.\text{Bond Length} \propto \frac{1}{\text{B.O.}}), the order of bond lengths is: O2+<O2<O2−<O22−\text{O}_2^+ < \text{O}_2 < \text{O}_2^- < \text{O}_2^{2-}

Hence, Statement (I) is True.

2. Analysis of Unpaired Electrons (Statement II):

From the electronic configurations:

  • O2\text{O}_2: Has 2 unpaired electrons in degenerate π2p∗\pi^*_{2p} orbitals (n=2n = 2).
  • O2+\text{O}_2^+: Has 1 unpaired electron in π2p∗\pi^*_{2p} orbital (n=1n = 1).
  • O2−\text{O}_2^-: Has 1 unpaired electron in π2p∗\pi^*_{2p} orbital (n=1n = 1).
  • O22−\text{O}_2^{2-}: Has 0 unpaired electrons as all orbitals are fully filled (n=0n = 0).

Therefore, the correct relation between the number of unpaired electrons is: O2 (2)>O2+ (1)=O2− (1)>O22− (0)\text{O}_2 \,(2) > \text{O}_2^+ \,(1) = \text{O}_2^- \,(1) > \text{O}_2^{2-} \,(0)

Since Statement (II) claims a strict inequality O2+>O2−\text{O}_2^+ > \text{O}_2^-, Statement (II) is False.

Conclusion:

  • Statement I is true.
  • Statement II is false.

This matches Option C.

Compare Bond Lengths and Unpaired Electrons in Dioxygen Species | Chemistry PYQ Solution - JEE Challenger